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# In the figure above, two security lights, L1 and L2 , are located 100

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Joined: 02 Sep 2009
Posts: 52253
In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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18 Feb 2018, 22:02
00:00

Difficulty:

65% (hard)

Question Stats:

46% (02:53) correct 54% (02:36) wrong based on 73 sessions

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In the figure above, two security lights, L1 and L2 , are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chain-link fence. What is the total length s of fence, in feet, illuminated by the two lights?

(A) 260

(B) 240

(C) 220

(D) 200

(E) 180

Attachment:

2018-02-19_1000.png [ 15.9 KiB | Viewed 911 times ]

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In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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18 Feb 2018, 23:35
Bunuel wrote:

In the figure above, two security lights, L1 and L2 , are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chain-link fence. What is the total length s of fence, in feet, illuminated by the two lights?

(A) 260

(B) 240

(C) 220

(D) 200

(E) 180

Attachment:
2018-02-19_1000.png

100 . 80 . 60 triangle is formed wth the radius, fence and distance between fence n light source

three such triangles i.e length = 80+80+80 = 240

but some length isnt convered by three triangles thus >240

(a) imo
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Re: In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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19 Feb 2018, 00:47

The lights make a 90 degree angle with the fence and hence the vertical distance is 60 feet. The lights cover 100 feet of radius , which means the triangle made is a pythagorus one with 100, 60 and uknown variable (calculate it to get 80). Thus, the light L1 covers a distance of 2(80) on fence and the L2 covers a little more than 80, making a total of more than 240.
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Re: In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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19 Feb 2018, 09:22
2
Attachment:

2018-02-19_1000.png [ 16.62 KiB | Viewed 709 times ]

The centers of both the circles are at a distance of 60 feet from the fence.

Since the maximum distance that the light will travel is 100 feet,
using the Pythagoras theorem the vertical distance can be calculated as $$\sqrt{100^2 - 60^2} = \sqrt{6400} = 80$$

But the small distance(marked in red) in the figure is not yet accounted for.
The total length(s) of the fence which is illuminated by the lights is a little over 240.

Therefore, Option A(260) is the only option available(> 240) and is our answer.
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Re: In the figure above, two security lights, L1 and L2 , are located 100  [#permalink]

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20 Apr 2018, 04:35
The distance from L1 to L2 is 100.
If we drop a perpendicular from L1 to the fence at P and from L2 to the fence at Q, then the distance between P & Q will be 100. (L1 L2 Q P will be a rectangle)
100 . 80 . 60 triangle is formed with the radius, fence and distance between fence & light source.
So total distance = 80+100+80 = 260
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Re: In the figure above, two security lights, L1 and L2 , are located 100 &nbs [#permalink] 20 Apr 2018, 04:35
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